Euclid’s Division Lemma

Step 1: (22) is a multiple of (11). Step 2: (11q+22=11(q+2)+0), so the correct remainder is (0). Step 3: If the added part is exactly divisible by the divisor, the remainder becomes zero.

Step 1: Divide (27) by (11). Step 2: (27=11 \times 2+5), so (11q+27=11(q+2)+5). Step 3: Finding the remainder of the large added part separately is an easy method.

Step 1: The number can be written as (16q+15). Step 2: The next number is (16q+16=16(q+1)+0). Step 3: After the greatest remainder, the next number has remainder (0).

Step 1: The number has the form (16q). Step 2: The next number is (16q+1), so the remainder is (1). Step 3: In consecutive numbers, remainders move in order like (0,1,2).

Step 1: In the Euclidean form (a=bq+r), (b) is the divisor. Step 2: In (618=47 \times 13+7), (47) is the number used for division. Step 3: Carefully identify the first number written in the product.

Step 1: In (a=bq+r), (q) is the quotient. Step 2: The number multiplied with (47) is (13), so (13) is the quotient. Step 3: To identify the quotient, look at the multiplier written with the divisor.

Step 1: In (a=bq+r), the final added part is (r). Step 2: Here (7) is less than (47), so it is the remainder. Step 3: While identifying the remainder, also check its range.

Step 1: Substitute in (a=bq+r): (93=5b+8). Step 2: (85=5b), so (b=17). Step 3: To find the unknown divisor, subtract the remainder first.

Step 1: Substitute in (a=bq+r): (108=15q+3). Step 2: (105=15q), so (q=7). Step 3: To find the unknown quotient, subtract the remainder first.

Step 1: The Euclidean division form is (a=bq+r). Step 2: Here the divisor is (14) and the remainder is (9), so the form is (14q+9). Step 3: In a general form, multiply the divisor by (q) and add the remainder.

Step 1: On division by (9), remainders can be from (0) to (8). Step 2: In (9q+9), the remainder is (9), which equals the divisor. Step 3: It should be written correctly as (9(q+1)).

Step 1: (33 \times 30=990). Step 2: (1000-990=10), so the remainder is (10). Step 3: For a large dividend, finding a nearby multiple saves time.

Step 1: (33 \times 30=990) and (33 \times 31=1023). Step 2: Since (1023) is greater, the quotient is (30). Step 3: The quotient is the greatest integer whose product with the divisor does not exceed the dividend.

Step 1: (2a=2(5q+4)=10q+8). Step 2: (8=5+3), so (2a=5(2q+1)+3) and the remainder is (3). Step 3: In multiplication-based questions, first multiply the old remainder and then divide by the divisor.

Step 1: The remainder of (a) is (5), so for (3a), check (3 \times 5=15). Step 2: (15=7 \times 2+1), so the remainder is (1). Step 3: In such questions, work with the remainder instead of the whole number.

Step 1: Write (a=4q+3). Step 2: (2a=8q+6=4(2q+1)+2), so the remainder is (2). Step 3: After multiplication, do not forget to convert the remainder into the correct range.

Step 1: Let (a=6q+5). Step 2: (2a=12q+10=6(2q+1)+4), so the remainder is (4). Step 3: (10) cannot be the remainder because it is greater than (6).

Step 1: The remainder is always less than the divisor. Step 2: The greatest integer less than (25) is (24). Step 3: The greatest remainder can be found quickly using (b-1).

Step 1: A remainder may start from (0). Step 2: If a number is exactly divisible by (25), the remainder is (0). Step 3: The smallest possible remainder is always (0).

Step 1: (39) is smaller than (100), so the quotient is (0). Step 2: (39=100 \times 0+39), and (39<100), so the form is correct. Step 3: If the dividend is smaller, the remainder can be the dividend itself.